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7 = -x+3. Consider the line y = Find the equation of the line that is parallel to this line and passes through the point (-5, 6). Find the equation of the line that is perpendicular to this line and passes through the point (-5, 6)

Respuesta :

Answer with explanation:

The equation of line is, y= -x +3

β†’x+y-3=0---------(1)

β‡’Equation of line Parallel to Line , ax +by +c=0 is given by, ax + by +K=0.

Equation of Line Parallel to Line 1 is

Β  x+y+k=0

The Line passes through , (-5,6).

β†’ -5+6+k=0

β†’ k+1=0

β†’k= -1

So, equation of Line Parallel to line 1 is

x+y-1=0

β‡’Equation of line Perpendicular Β to Line , ax +by +c=0 is given by, bx - a y +K=0.

Equation of Line Perpendicular to Line 1 is

Β  x-y+k=0

The Line passes through , (-5,6).

β†’ -5-6+k=0

β†’ k-11=0

β†’k= 11

So, equation of Line Parallel to line 1 is

x-y+11=0

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